Crack problems concerning boundaries of convex lens like forms

International Journal of Trend in Scientific Research and Development(IJTSRD)
ISSN:2456-6470 —www.ijtsrd.com—Volume -2—Issue-3
CRACK PROBLEMS CONCERNING BOUNDARIES OF CONVEX LENS LIKE
FORMS
DOO-SUNG LEE
Department of Mathematics
College of Education, Konkuk University
120 Neungdong-Ro, Kwangjin-Gu, Seoul, Korea
e-mail address: dslee@kumail.konkuk.ac.kr
Abstract
The singular stress problem of a peripheral edge
crack around a cavity of spherical portion in an
infinite elastic medium when the crack is sub-
jected to a known pressure is investigated. The
problem is solved by using integral transforms
and is reduced to the solution of a singular in-
tegral equation of the first kind. The solution
of this equation is obtained numerically by the
method due to Erdogan, Gupta , and Cook, and
the stress intensity factors are displayed graphi-
cally.
Also investigated in this paper is the penny-
shaped crack situated symmetrically on the cen-
tral plane of a convex lens shaped elastic mate-
rial.
Key words: cavity of spherical portion/ periph-
eral edge crack/penny-shaped crack /SIF.
1.Introduction.
The problem of determining the distribution
of stress in an elastic medium containing a cir-
cumferential edge crack has been investigated by
several researchers including the present author.
Among these investigations, the notable ones
are Keer et al.[1,2],Atsumi and Shindo[3,4], and
Lee[5,6]. Keer et al.[1] considered a circumfer-
ential edge crack in an extended elastic medium
with a cylindrical cavity the analysis of which
provides immediate application to the study of
cracking of pipes and nozzles if the crack is small.
Another important problem involving a cir-
cumferential edge crack is that concerned with a
spherical cavity. Atsumi and Shindo[4] investi-
gated the singular stress problem of a peripheral
edge crack around a spherical cavity under uni-
axial tension field. In more recent years, Wan
et al.[7] obtained the solution for cracks emanat-
ing from surface semi-spherical cavity in finite
body using energy release rate theory. In previ-
ous studies concerning the spherical cavity with
the circumferential edge crack, the cavity was a
full spherical shape. In this present analysis, we
are concerned with a cavity of a spherical por-
tion, rather than a full spherical cavity. More
briefly describing it, the cavity looks like a con-
vex lens. Here, we employ the known methods of
previous investigators to derive a singular inte-
gral equation of the first kind which was solved
numerically, and obtained the s.i.f. for various
spherical portions. It is also shown that when
this spherical portion becomes a full sphere, the
present solution completely agrees with the al-
ready known solution.
2.Formulation of problem and reduction to
singular integral equation. 0
We employ cylindrical coordinates (r, φ, z)
with the plane z = 0 coinciding the plane of pe-
ripheral edge crack. The spherical coordinates
0
@IJTSRD— Available Online@www.ijtsrd.com—Volume-2—Issue-3—Mar-Apr 2018 Page:982
1

2INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-6470
(ρ, θ, φ) are connected with the cylindrical coor-
dinates by
z = ρ cos θ, r = ρ sin θ.
Spherical coordinates (ζ, ϑ, φ) whose origin is
located at z = −δ, r = 0, and is the center of
the upper spherical surface, are also used. The
cavity is symmetrical with respect to the plane
z = 0.
The crack occupies the region z = 0, 1 ≤
r ≤ γ. So the radius of the spherical cavity is
ζ0 =
√
1 + δ2.
The boundary conditions are:
On the plane z = 0, we want the continuity of
the shear stress, and the normal displacement:
uz(r, 0+
)−uz(r, 0−
) = 0, γ ≤ r < ∞, (2.1)
σrz(r, 0+
) − σrz(r, 0−
) = 0, 1 ≤ r < ∞.
(2.2)
And the crack is subjected to a known pressure
p(r), i.e.,
σzz(r, 0+
) = −p(r), 1 ≤ r ≤ γ. (2.3)
On the surface of the spherical cavity, stresses
are zero:
σζζ(ζ0, ϑ) = 0, (2.4)
σζϑ(ζ0, ϑ) = 0. (2.5)
We can make use of the axially symmetric so-
lution of the equations of elastic equilibrium due
to Green and Zerna [8] which states that if ϕ(r, z)
and ψ(r, z) are axisymmetric solutions of Laplace
equation, then the equations
2µur =
∂ϕ
∂r
+ z
∂ψ
∂r
, (2.6)
2µuz =
∂ϕ
∂z
+ z
∂ψ
∂z
− (3 − 4ν)ψ, (2.7)
where µ is the modulus of rigidity and ν is Pois-
son’s ratio, provide a possible displacement ﬁeld.
The needed components of stress tensor are given
by the equations
σrz =
∂2φ
∂r∂z
+ z
∂2ψ
∂r∂z
− (1 − 2ν)
∂ψ
∂r
, (2.8)
σzz =
∂2φ
∂z2
+ z
∂2ψ
∂z2
− 2(1 − ν)
∂ψ
∂z
. (2.9)
The functions φ(1) and φ(2) for the regions z > 0
and z < 0, respectively, are chosen as follows:
φ(1)
(r, z) = (2ν − 1)
∞
0
ξ−1
A(ξ)J0(ξr)e−ξz
dξ
+
∞
n=0
an
Pn(cos θ)
ρn+1
, (2.10)
φ(2)
(r, z) = (2ν − 1)
∞
0
ξ−1
A(ξ)J0(ξr)eξz
dξ
−
∞
n=0
an(−1)n Pn(cos θ)
ρn+1
. (2.11)
Here the superscripts (1) and (2) are taken for
the region z > 0 and z < 0, respectively. The
functions ψ(1) and ψ(2) are chosen as follows:
ψ(1)
(r, z) =
∞
0
A(ξ)J0(ξr)e−ξz
dξ
+
∞
n=0
bn
Pn(cos θ)
ρn+1
, (2.12)
ψ(2)
(r, z) = −
∞
0
A(ξ)J0(ξr)eξz
dξ
+
∞
n=0
bn(−1)n Pn(cos θ)
ρn+1
. (2.13)
Then we can immediately satisfy condition (2.2)
by this choice of functions (2.10)-(2.13).
Now the condition (2.1) requires
∞
0
A(ξ)J0(ξr)dξ = 0, r > γ. (2.14)
0
Equation (2.14) is automatically satisﬁed by
setting
A(ξ) =
γ
1
tg(t)J1(ξt)dt. (2.15)
0@IJTSRD— Available Online@www.ijtsrd.com—Volume-2—Issue-3—Mar-Apr 2018 Page:983

INTERNATIONAL JOURNAL OF TREND IN SCIENTIFIC RESEARCH AND DEVELOPMENT(IJTSRD)ISSN:2456-64703
Then from the boundary condition (2.3), we
obtain
∞
0
ξA(ξ)J0(ξr)dξ −
∞
n=0
a2n
(2n + 1)2P2n(0)
r2n+3
−2(1 − ν)
∞
n=0
b2n+1
P2n+1(0)
r2n+3
= −p(r),
1 ≤ r ≤ γ, (2.16)
where prime indicates the differentiation with re-
spect to the argument.
By substituting (2.15) into (2.16), it reduces
to
−
2
π
γ
1
tg(t)R(r, t)dt −
∞
n=0
r−(2n+3)
{P2n(0)
×(2n + 1)2
a2n + αb2n+1P2n+1(0)} = −p(r),
1 ≤ r ≤ γ, (2.17)
where
R(r, t) =
1
r2 − t2
E
r
t
, t > r,
=
r
t
1
r2 − t2
E
t
r
−
1
rt
K
t
r
, r > t.
(2.18)
K and E in (2.18) are complete elliptic integrals
of the first and the second kind, respectively and
α = 2(1 − ν).
The solution will be complete, if the condi-
tions on the surface of the spherical cavity are
satisfied.
3.Conditions on the surface of the spheri-
cal cavity.
Equation (2.17) gives one relation connecting
unknown coefficients an and bn. The stress com-
ponents besides (2.8) and (2.9) which are needed
for the present analysis are given by the following
equations
σζζ =
∂2φ
∂ζ2
+ ζ cos ϑ
∂2ψ
∂ζ2
− 2(1 − ν) cos ϑ
∂ψ
∂ζ
+2ν
sin ϑ
ζ
∂ψ
∂ϑ
, (3.1)
σζϑ =
1
ζ
∂2φ
∂ζ∂ϑ
−
1
ζ2
∂φ
∂ϑ
+ cos ϑ
∂2ψ
∂ζ∂ϑ
+(1 − 2ν) sin ϑ
∂ψ
∂ζ
− 2(1 − ν)
cos ϑ
ζ
∂ψ
∂ϑ
. (3.2)
To satisfy boundary conditions on the spherical
surface, it is needed to represent φ, ψ in (2.10)-
(2.13) in terms of ζ, ϑ variables. To do so we uti-
lize the following formula whose validity is shown
in the Appendix 1. An expression useful for the
present analysis is the following
Pn(cos θ)
ρn+1
=
∞
k=0
n + k
k
Pn+k(cos ϑ)
ζn+k+1
δk
.
(3.3)
Thus
∞
n=0
an
Pn(cos θ)
ρn+1
=
∞
n=0
an
∞
k=0
n + k
k
×
Pn+k(cos ϑ)
ζn+k+1
δk
=
∞
j=0
Pj(cos ϑ)
ζj+1
Aj, (3.4)
where
Aj =
j
n=0
j!
(j − n)!n!
anδj−n
. (3.5)
Also
∞
0
ξ−1
A(ξ)J0(ξr)e−ξz
dξ =
γ
1
tg(t)
×
∞
0
ξ−1
J0(ξr)J1(ξt)e−ξz
dξdt. (3.6)
If we make use of the formula in Whittaker and
Watson[9,pp.395-396]
π
−π
exp{−ξ(z+ix cos u+iy sin u)}du = 2πe−ξz
J0(ξr),
0
to the inner integral of (3.6), it can be written
as, if we are using the shortened notation
β = z + ix cos u + iy sin u,
then
∞
0
ξ−1
dξ
=
1
2π
π
−π
∞
0
ξ−1
J1(ξt)e−ξβ
dξdu

=
1
2π
π
−π
t
β + β2 + t2
du
= −
1
2π
π
−π
β
t
− 1 +
β2
t2
du
= −
1
2π
π
−π
β
t
−
∞
n=0
(−1
2 )n(−1)n
n!
β
t
2n
du.
(3.7)
As |β| < |t|, the above series expansion is valid.
Next,
β2n
= (−δ+Z+ix cos u+iy sin u)2n
= (−δ+β )2n
=
2n
k=0
2n
k
(−δ)2n−k
(β )k
, (3.8)
where (r, φ, Z) is the cylindrical coordinates sys-
tem centered at (r, z) = (0, −δ). We have
the following formula from Whittaker and Wat-
son[9,p.392]
π
−π
(β )n
du =
π
−π
(Z + ix cos u + iy sin u)n
du
= 2πζn
Pn(cos ϑ). (3.9)
If we utilize (3.8) and (3.9) in (3.7), it can be
written as
∞
0
ξ−1
dξ
=
∞
n=0
(−1
2 )n(−1)n
n!
1
t2n
×
2n
k=0
2n
k
(−δ)2n−k
ζk
Pn(cos ϑ)
+δ −
ζ
t
P1(cos ϑ). (3.10)
Thus finally, from (3.4), (3.6) and (3.10), φ(1)
can be written in terms of spherical coordinates
(ζ, ϑ) as
φ(1)
=
∞
k=0
Pk(cos ϑ)
Ak
ζk+1
+ Φkζk
+δ
γ
1
tg(t)dt − ζP1(cos ϑ)
γ
1
g(t)dt,
where
Φk = −(α − 1)
∞
n=[(k+1)/2]
(−1
2 )n(−1)n
n!
×
(2n)!
(2n − k)!k!
(−δ)2n−k
γ
1
g(t)
t2n−1
dt, (3.11)
and [(k+1)/2] is the greatest integer ≤ (k+1)/2.
It is also necessary to express ψ(1) in terms of
spherical coordinates (ζ, ϑ). Now, as in (3.4)
∞
n=0
bn
Pn(cos θ)
ρn+1
=
∞
j=0
Pj(cos ϑ)
ζj+1
Bj, (3.12)
where
Bj =
∞
n=0
j!
(j − n)!n!
bnδj−n
.
We first express following integral in (ρ, θ) co-
ordinates
∞
0
A(ξ)J0(ξr)e−ξz
dξ =
γ
1
tg(t)dt
×
∞
0
J1(ξt)J0(ξr)e−ξz
dξ. (3.13)
The inner integral on the right-hand side of
(3.13) is
−
1
t
∞
0
J0(ξr)e−ξz ∂
∂ξ
J0(ξt)dξ =
1
t
−
1
t
∞
0
{rJ1(ξr) + zJ0(ξr)}e−ξz
J0(ξt)dξ.
(3.14)
Using the equation in Erdélyi et al.[10] 0
J0(ξt) =
2
π
∞
t
sin(ξx)dx
√
x2 − t2
,
equation (3.14) is equal to
1
t
+
1
t
2
π
∞
t
dx
√
x2 − t2
×
∞
0
{rJ1(ξr) + zJ0(ξr)}e−ξ(z+ix)
dξ
=
1
t
+
1
t
2
π
∞
t
1
√
x2 − t2
−ix
r2 + (z + ix)2
dx
=
1
t
−
1
t
2
π
∞
t
1
√
x2 − t2
dx
1 − 2ρ
xi cos θ + (ρ
x i)2
=
1
t
−
1
t
∞
n=0
ρ
t
2n+1
(−1)n
P2n+1(cos θ)
(1
2 )n
n!
,
(3.15)

where we used the generating function
1
√
1 − 2x cos θ + x2
=
∞
n=0
Pn(cos θ)xn
.
To express (3.15) in the (ζ, ϑ) spherical coordi-
nates, we use the following formula whose valid-
ity is shown in the Appendix 2.
ρ2n+1
P2n+1(cos θ) =
2n+1
k=0
(2n + 1)!
k!(2n + 1 − k)!
×(−δ)2n+1−k
ζk
Pk(cos ϑ). (3.16)
0
If we use (3.16) in (3.15), then with (3.12), we
get ψ(1) in spherical coordinates (ζ, ϑ)
ψ(1)
=
∞
k=0
Pk(cos ϑ)
Bk
ζk+1
+Ψkζk
+
γ
1
g(t)dt,
(3.17)
where
Ψk = −
∞
n=[k/2]
(−1)n(1
2 )n
n!
(2n + 1)!
k!(2n + 1 − k)!
×(−δ)2n+1−k
γ
1
g(t)
t2n+1
dt.
Then if we substitute these values of φ(1) and
ψ(1) given by (3.11) and (3.17) into (3.2), and
simplify the results by using the properties of
Legendre polynomials, from the condition that
on the spherical surface ζ = ζ0, the shear stress
σζϑ = 0, we obtain the following equation
−Ak+1
k + 3
ζk+4
0
+
Bk
ζk+2
0
α − (k + 1)2
2k + 1
−
Bk+2
ζk+4
0
(k + 3)(2α + k + 2)
2k + 5
+kζk−1
0 Φk+1 + ζk−1
0 Ψk
k(k + 1 − 2α)
2k + 1
−ζk+1
0 Ψk+2
k + 2 + α − (k + 2)(k + 3)
2k + 5
= 0.
(3.18)
Likewise, from the condition σζζ(ζ0, ϑ) = 0,
we get following two equations,
−A0
2
ζ3
0
−
B1
ζ3
0
2(2α + 1)
3
−
1
3
Ψ1(α − 4) = 0,
(3.19a)
−Ak+1
(k + 2)(k + 3)
ζk+4
0
+
Bk
ζk+2
0
(k + 1){2 − α − (k + 1)(k + 4)}
2k + 1
−
Bk+2
ζk+4
0
(k + 2)(k + 3)(2α + k + 2)
2k + 5
−k(k+1)ζk−1
0 Φk+1−ζk−1
0 Ψk
k(k + 1)(k + 1 − 2α)
2k + 1
−ζk+1
0 Ψk+2
(k + 2){(k + 2)(k − 1) + α − 2}
2k + 5
= 0.
(3.19b)
Thus if we multiply (3.18) by k + 2 and subtract
(3.19b) from the resulting equation we ﬁnd
Bk = −
ζk+2
0 (2k + 1)
α(2k + 3) + 2k(k + 1)
k(2k+3)ζk−1
0 Φk+1
+ζk−1
0 Ψk
(2k + 3)k(k + 1 − 2α)
2k + 1
+ζk+1
0 Ψk+2k(k+2) .
(3.20)
If we solve (3.20) for bi using the theorem
which is in the Appendix 3 and the relation,
Φk+1 =
α − 1
k + 1
Ψk,
we ﬁnd
bi
δi
=
i
k=0
i!(−1)i−k
(i − k)!k!
1
δk
[N1(k)Ψk + N2(k)Ψk+2],
(3.21)
where
N1(k) = −
ζ2k+1
0 (2k + 3)k
α(2k + 3) + 2k(k + 1)
−α + k2
k + 1
,
N2(k) = −
ζ2k+3
0 (2k + 1)k(k + 2)
α(2k + 3) + 2k(k + 1)
.
Equation (3.21) can be written as
bi = −
i
k=0
i!(−1)i−k
(i − k)!k!
δi
δk
γ
1
g(t)
t
×
1
n=0
[Nn+1(k)fk+2n
1
2
,
δ
t
dt, (3.22)

where
fk(c, x) =
(−δ)1−k
k!
∞
=[k/2]
(−1) (c) (2 + 1)!x2
(2 + 1 − k)! !
with (c) = c(c + 1)(c + 2) · · · (c + − 1).
Now from (3.19a,b) we have
Ak+1 = M1(k)Ψk + M2(k)Ψk+2 + M3(k)Ψk+4,
A0 = Ã1Ψ1 + Ã2Ψ3,
where 0
M1(k) = −
ζ2k+3
0 k(−α + k2)
(k + 2)(k + 3)(2k + 1)
×
(2k + 3)˜g(k)
h(k)
+ 1 ,
M2(k) =
ζ2k+5
0
k + 3
−
˜g(k)
h(k)
k(k+1)+
˜g(k − 2)
2k + 5
+
(k + 2)(2k + 7)(2α + k + 2){−α + (k + 2)2}
(2k + 5)h(k + 2)
,
M3(k) =
ζ2k+7
0 (k + 2)(k + 4)(2α + k + 2)
h(k + 2)
,
Ã1 = −
ζ3
0
6
α − 4 −
5(2α + 1)(−α + 1)
h(1)
,
Ã2 =
3(2α + 1)ζ5
0
h(1)
,
with
˜g(k) = 2 − α − (k + 1)(k + 4),
h(k) = α(2k + 3) + 2k(k + 1).
In a similar way ais can be found from these
equations as follows:
a2i = ( Ã1Ψ1 + Ã2Ψ3)δ2i
+
2i−1
k=0
(2i)!(−δ)2i−k−1
(2i − 1 − k)!(k + 1)!
˜F(k),
where
˜F(k) = −
γ
1
g(t)
t
M1(k)fk
1
2
,
δ
t
+ M2(k)
×fk+2
1
2
,
δ
t
+ M3(k)fk+4
1
2
,
δ
t
dt.
Thus
∞
i=0
a2i
(2i + 1)2P2i(0)
r2i+3
=
a0
r3
+
∞
i=1
( Ã1Ψ1 + Ã2Ψ3)
×δ2i (2i + 1)2(−1)i(1
2)i
r2i+3i!
+
∞
i=1
(2i + 1)2(−1)i(1
2 )i
r2i+3i!
×
2i−1
k=0
(2i)!(−δ)2i−k−1
(2i − 1 − k)!(k + 1)!
˜F(k).
If we interchange the order of summation in the
last term of the above equation, and use
Ψ1 = −
γ
1
g(t)
t
f1
1
2
,
δ
t
dt,
Ψ3 = −
γ
1
g(t)
t
f3
1
2
,
δ
t
dt,
we finally obtain
∞
i=0
a2i
(2i + 1)2P2i(0)
r2i+3
= −
γ
1
tg(t)T1(r, t)dt,
where
T1(r, t) =
1
r3t2
×
∞
k=0
2
m=0
Mm+1(k)fk+2m
1
2
,
δ
t
hk
δ
r
+ Ã1f1
1
2
,
δ
t
+ Ã2f3
1
2
,
δ
t
1 + j
δ
r
,
with
hk
δ
r
=
(−δ)−k−1
(k + 1)!
×
∞
i=[k/2+1]
(2i + 1)2(−1)i(1
2 )i(2i)!
(2i − 1 − k)!i!
δ
r
2i
,
j
δ
r
=
∞
i=1
(2i + 1)2(−1)i(1
2 )i
i!
δ
r
2i
.
Also,
∞
i=0
b2i+1
P2i+1(0)
r2i+3
= −
∞
i=0
(−1)i(3
2 )i
i!r2i+3
×
2i+1
k=0
(2i + 1)!(−δ)2i−k+1
(2i + 1 − k)!k!

×
γ
1
g(t)
t
1
n=0
Nn+1(k)fk+2n
1
2
,
δ
t
dt.
If we change the order of summation in the above
equation, we get
∞
i=0
b2i+1
P2i+1(0)
r2i+3
= −
γ
1
tg(t)T2(r, t)dt,
where
T2(r, t) =
1
t2r3
∞
k=0
1
n=0
Nn+1(k)fk
3
2
,
δ
r
×fk+2n
1
2
,
δ
t
.
Thus ﬁnally the singular integral equation (2.17)
becomes
2
π
γ
1
tg(t){R(r, t) + S(r, t)}dt = p(r),
1 ≤ r ≤ γ, (3.23)
where
S(r, t) = −
π
2
{T1(r, t) + αT2(r, t)}.
When δ = 0, we brieﬂy show that this equation
completely agrees with what Atsumi and Shindo
obtained. Using
lim
δ→0
h2k−1
δ
r
=
(−1)k(1
2 )k(2k + 1)2
k!
1
r2k
,
lim
δ→0
f2k−1
1
2
,
δ
t
=
(−1)k−1(1
2 )k−1
(k − 1)!
1
t2k−2
,
A = lim
δ→0
∞
k=0
2
m=0
Mm+1(k)fk+2m
1
2
,
δ
t
hk
δ
r
=
∞
k=1
(−1)k(1
2)k(2k + 1)2
k!r2k
×
1
(2k + 1)(2k + 2)
2k − 1
4k − 1
(4k + 1)G(k)
H(k)
+ 1
×F(k)
(−1)k−1(1
2 )k−1
(k − 1)!
1
t2k−2
+(2k + 1) −
G(k)
H(k)
(2k − 1)2k
−
(2k + 1)(4k + 5)(2α + 2k + 1)
H(k + 1)(4k + 3)
F(k + 1)
+
G(k − 1)
4k + 3
(−1)k(1
2 )k
k!
1
t2k
+
(2k + 1)2(2k + 2)(2k + 3)(2α + 2k + 1)
H(k + 1)
×
(−1)k+1(1
2 )k+1
(k + 1)!
1
t2k+2
,
where 0
H(k) = α(4k + 1) + 4k(2k − 1),
G(k) = 2 − α − 2k(2k + 3),
F(k) = −α + (2k − 1)2
.
B = lim
δ→0
∞
k=0
1
n=0
Nn+1(k)fk
3
2
,
δ
r
fk+2n
1
2
,
δ
t
=
∞
k=1
(−1)k(3
2 )k
k!r2k
−
(2k + 1)(4k + 5)
2(k + 1)
×
F(k + 1)
H(k + 1)
(−1)k(1
2 )k
k!
1
t2k
−
(4k + 3)(2k + 1)(2k + 3)
H(k + 1)
×
(−1)k+1(1
2 )k+1
(k + 1)!
1
t2k+2
+
1
2H(1)
−5F(1)+
9
t2
.
Thus
A + αB =
∞
k=1
(−1)k(1
2 )k(2k + 1)
k!r2k
×
(−1)k(2k − 1)(2k + 1)
H(k)k(k + 1)
k(k + 1)F(k)
−
1
t2
1
4(4k + 3)
{(2k − 1)(4k + 3)2kG(k)
−H(k)G(k − 1)}
(1
2)k−1
(k − 1)!
1
t2k−2
+(−1)k F(k + 1)(2k + 1)
H(k + 1)(2k + 2)
(2k + 3)(2k + 2)
1
t2
+F(k+1)
4k + 5
4k + 3
(1
2 )k
k!
1
t2k
+
α
2H(1)
9
t2
−5F(1)
=
∞
k=1
2(2k + 1)2
H(k)(k + 1)
k(k + 1)F(k) −
1
t2
1
4(4k + 3)
×{(2k − 1)(4k + 3)2kG(k) − H(k)G(k − 1)}
×
1
(rt)2k−2r2
(1
2 )k
k!
2
+
∞
k=2
F(k)
H(k)
(2k + 1)(2k − 1)
1
t2
+ F(k)
4k + 1
4k − 1

×2k
(1
2 )k
k!
2
1
(rt)2k−2
+
α
2H(1)
−5F(1) +
9
t2
.
Using
Ã1f1
1
2
, 0 + Ã2f3
1
2
, 0 = Ã1 − Ã2
1
2t2
,
we finally obtain
−
2
π
S(r, t) =
1 + ν
3
1
r3t2
+
∞
k=1
2
H(k)
2(2k + 1)2
(k + 1)r2
× k(k + 1)F(k) −
1
t2
1
4(4k + 3)
{(2k − 1)(4k + 3)
×2kG(k) − H(k)G(k − 1)} + kF(k) (2k + 2)
×(2k − 1)
1
t2
+ F(k)
4k + 1
4k − 1
×
(2k − 1)!!
(2k)!!
2
1
(rt)2kr
.
The above equation completely agrees with the
result by Atsumi and Shindo.
A quantity of physical interest is the stress in-
tensity factor which is given as
K = lim
r→γ+
2(r − γ)σzz(r, 0).
We choose p(r) = p0 and put γ − 1 = a. We let
r =
a
2
(s + 1) + 1, t =
a
2
(τ + 1) + 1, (3.24)
and in order to facilitate numerical analysis, as-
sume g(t) to have the following form:
g(t) = p0(t − 1)
1
2 (γ − t)−1
2 ˜G(t). (3.25)
With the aid of (3.24), g(τ) can be rewritten as
g(τ) = p0
˜G(τ)
1 + τ
1 − τ
1
2
. (3.26)
The stress intensity factors K can therefore be
expressed in terms of ˜G(t) as
K/p0 =
√
2a ˜G(γ), (3.27)
or in terms of the quantity actually calculated
K/p0
√
a =
√
2 ˜G(γ). (3.28)
4.Numerical analysis.
In order to obtain numerical solution of (3.23),
substitutions are made by the application of
(3.24) and (3.26) to obtain the following expres-
sion:
a
π
1
−1
1 + τ
1 − τ
1
2
˜G(τ)
a
2
(τ + 1) + 1 [R(s, τ)
+S(s, τ)]dτ = 1, −1 < s < 1. (4.1)
The numerical solution technique is based on
the collocation scheme for the solution of singu-
lar integral equations given by Erdogan, Gupta,
and Cook [11]. 0
This amounts to applying a
Gaussian quadrature formula for approximating
the integral of a function f(τ) with weight func-
tion [(1+τ)/(1−τ)]
1
2 on the interval [-1,1]. Thus,
letting n be the number of quadrature points,
1
−1
1 + τ
1 − τ
1
2
f(τ)dτ
2π
2n + 1
n
k=1
(1 + τk)f(τk),
(4.2)
where
τk = cos
2k − 1
2n + 1
π, k = 1, . . . n. (4.3)
The solution of the integral equation is ob-
tained by choosing the collocation points:
si = cos
2iπ
2n + 1
, i = 1, . . . , n, (4.4)
and solving the matrix system for G∗(τk) :
n
k=1
[R(sj, τk) + S(sj, τk)]G∗
(τk) =
2n + 1
2a
,
j = 1, . . . , n, (4.5)
where
˜G(τk) =
G∗(τk)
(1 + τk)[a
2 (τk + 1) + 1]
. (4.6)
5.Numerical results and consideration.
Numerical calculations have been carried out
for ν = 0.3. The values of normalized stress
intensity factor K/p0
√
a versus a are shown in
Fig.1-3 for various values of δ.
Fig.1 shows the variation of K/p0
√
a with re-
spect to a when δ = 0.
This figure shows that as a increases, S.I.F.
decreases steadily.

Fig.2 and 3 deal with the cases when δ = 0.3
and δ = 0.5, respectively. Here we omit units.
We can see that the trend is similar. Theoreti-
cally the inﬁnite series involved converges when
δ < 1 by comparison test. However, because of
the overﬂow, computations could not be accom-
plished beyond the values δ > 0.5. And we found
that the variation of SIF is very small with re-
spect to the variation of δ.
0

6.Penny-shaped crack.
In this section we are concerned with a penny-
shaped crack in a convex lens shaped elastic ma-
terial. The problem of determining the distri-
bution of stress in an elastic sphere containing
a penny-shaped crack or the mixed boundary
value problems concerning a spherical boundary
has been investigated by several researchers. Sri-
vastava and Dwivedi [12] considered the prob-
lem of a penny-shaped crack in an elastic sphere,
whereas Dhaliwal et al.[13] solved the problem of
a penny-shaped crack in a sphere embedded in an
inﬁnite medium. On the other hand, Srivastav
and Narain [14] investigated the mixed boundary
value problem of torsion of a hemisphere.
7.Formulation of problem and reduction to
a Fredholm integral equation of the sec-
ond kind. 0
We employ cylindrical coordinates
(r, φ, z) with the plane z = 0 coinciding the plane
of the penny shaped crack. The center of the
crack is located at (r,z)=(0,0). As before, spher-
ical coordinates (ρ, θ, φ) are connected with the
cylindrical coordinates by
z = ρ cos θ, r = ρ sin θ.
Spherical coordinates (ζ, ϑ, φ) whose origin is at
z = −δ, r = 0, and is the center of the upper
spherical surface of the convex elastic body, is
also used. The elastic body is symmetrical with
respect to the plane z = 0.
The crack occupies the region z = 0, 0 ≤
r ≤ 1. The radius of the spherical portion is
ζ0 = γ2 + δ2. The boundary conditions are:
On the plane z = 0, we want the continuity of
the shear stress, and the normal displacement:
uz(r, 0+
) − uz(r, 0−
) = 0, 1 ≤ r ≤ γ, (7.1)
σrz(r, 0+
)−σrz(r, 0−
) = 0, 0 ≤ r ≤ γ. (7.2)
And the crack is subjected to a known pressure
p(r), i.e.,
σzz(r, 0+
) = −p(r), 0 ≤ r ≤ 1. (7.3)
On the surface of the spherical portion, stresses
are zero:
σζζ(ζ0, ϑ) = 0, (7.4)
σζϑ(ζ0, ϑ) = 0. (7.5)
We can make use of (2.6)-(2.9) also, for the
present case. The functions φ(1) and φ(2) for the
regions z > 0 and z < 0, respectively, are chosen
as follows:
φ(1)
(r, z) = (2ν − 1)
∞
0
ξ−1
A(ξ)J0(ξr)e−ξz
dξ

+
∞
n=0
anρn
Pn(cos θ), (7.6)
φ(2)
(r, z) = (2ν − 1)
∞
0
ξ−1
A(ξ)J0(ξr)eξz
dξ
−
∞
n=0
an(−1)n
ρn
Pn(cos θ). (7.7)
Here the superscripts (1) and (2) are taken for
the region z > 0 and z < 0, respectively. The
functions ψ(1) and ψ(2) are chosen as follows:
ψ(1)
(r, z) =
∞
0
A(ξ)J0(ξr)e−ξz
dξ
+
∞
n=0
bnρn
Pn(cos θ), (7.8)
ψ(2)
(r, z) = −
∞
0
A(ξ)J0(ξr)eξz
dξ
+
∞
n=0
bn(−1)n
ρn
Pn(cos θ). (7.9)
Then we can immediately satisfy condition (7.2)
by these choice of functions (7.6)-(7.9).
Now the condition (7.1) requires
∞
0
A(ξ)J0(ξr)dξ = 0, r > 1. (7.10)
Equation (7.10) is automatically satisfied by
setting
A(ξ) =
1
0
g(t) sin(ξt)dt, g(0) = 0. (7.11)
Then from the boundary condition (7.3), we ob-
tain
∞
0
ξA(ξ)J0(ξr)dξ −
∞
n=0
a2n(2n)2
P2n(0)r2n−2
−2(1 − ν)
∞
n=0
b2n+1P2n+1(0)r2n
= −p(r),
0 ≤ r ≤ 1, (7.12)
where the prime indicates the differentiation
with respect to the argument. 0
By substituting (7.11) into (7.12), it reduces
to
g(t) −
2
π
∞
n=0
(−1)n
{2nt2n−1
a2n
+αb2n+1t2n+1
} = h(t), 0 ≤ r ≤ 1, (7.13)
where
h(t) = −
2
π
t
0
rp(r)
√
t2 − r2
dr.
The solution will be complete, if the condi-
tions on the surface of the spherical portion are
satisfied.
8.Conditions on the surface of the sphere.
Equation (7.13) gives one relation connecting
unknown coefficients an and bn. The stress com-
ponents besides (2.8) and (2.9) which are needed
for the present analysis are given by (3.1) and
(3.2). To satisfy boundary conditions on the
spherical surface, it is needed to represent φ, ψ
in (7.6)-(7.9) in terms of ζ, ϑ variables. To do so
we utilize the following formula in the Appendix
2. An expression useful for the present analysis
is the following
Pn(cos θ)ρn
=
n
k=0
n
k
Pk(cos ϑ)ζk
(−δ)n−k
.
(8.1)
Thus
∞
n=0
anPn(cos θ)ρn
=
∞
n=0
an
∞
k=0
n
k
Pk(cos ϑ)
×ζk
(−δ)n−k
=
∞
j=0
Pj(cos ϑ)ζj
Aj, (8.2)
where
Aj =
∞
n=j
n!
(n − j)!j!
an(−δ)n−j
. (8.3)
Also
−
∂
∂t
∞
0
ξ−1
A(ξ)J0(ξr)e−ξz
dξ
= −
1
0
g(t)
∞
0
J0(ξr) cos(ξt)e−ξz
dξdt. (8.4)

The inner integral of the above equation is
∞
0
J0(ξr) cos(ξt)e−ξz
dξ
=
∞
0
J0(ξr)e−ξ(z+it)
dξ
=
1
r2 + (z + it)2
=
1
√
r2 + z2 + 2zit − t2
=
1
ρ 1 − 2 cos θ(−it
ρ ) + (−it
ρ )2
=
1
ρ
∞
n=0
t
ρ
2n
(−1)n
P2n(cos θ). (8.5)
Thus ﬁnally, from (8.2), and using the for-
mula in Appendix, φ(1) can be written in terms
of spherical coordinates (ζ, ϑ) as
φ(1)
=
∞
k=0
Pk(cos ϑ)
Φk
ζk+1
+ Akζk
, (8.6)
where
Φk = −(α − 1)
[k/2]
n=0
k!
(2n)!(k − 2n)!
δk−2n
×(−1)n
1
0
g(t)t2n+1
2n + 1
dt. (8.7)
It is also necessary to express ψ(1) in terms of
spherical coordinates (ζ, ϑ). Now, as in (8.2)
∞
n=0
bnPn(cos θ)ρn
=
∞
j=0
Pj(cos ϑ)ζj
Bj, (8.8)
where
Bj =
∞
n=j
n!
j!(n − j)!
bn(−δ)n−j
. (8.9)
0
We ﬁrst express following integral in (ρ, θ) co-
ordinates
∞
0
A(ξ)J0(ξr)e−ξz
dξ =
1
0
g(t)dt
×
∞
0
sin(ξt)J0(ξr)e−ξz
dξ. (8.10)
The inner integral on the right-hand side of
(8.10) is
−
∞
0
J0(ξr)e−ξ(z+it)
dξ
=
∞
n=0
P2n+1(cos θ)
ρ2n+2
(−1)n
t2n+1
. (8.11)
To express (8.11) in the (ζ, ϑ) spherical co-
ordinates, we use the following formula in the
Appendix 1.
P2n+1(cos θ)
ρ2n+1
=
∞
k=0
(2n + 1 + k)!
k!(2n + 1)!
×δk P2n+1+k(cos ϑ)
ζ2n+2+k
. (8.12)
If we use (8.8) and (8.12), we get ψ(1) in spherical
coordinates (ζ, ϑ) as
ψ(1)
=
∞
k=0
Pk(cos ϑ)
Ψk
ζk+1
+ Bkζk
+ B0,
(8.13)
where
Ψk =
[(k−1)/2]
n=0
k!
(2n + 1)!(k − 2n − 1)!
δk−2n−1
×(−1)n
1
0
g(t)t2n+1
dt. (8.14)
Then if we substitute these values of φ(1) and
ψ(1) given by (8.6) and (8.13) into (3.2), and
simplify the results by using the properties of
Legendre polynomials, from the condition that
on the spherical surface ζ = ζ0, σζϑ = 0, we
obtain the following equation
Ak+1kζk−1
0 − Bk+2ζk+1
0
α − (k + 2)2
2k + 5
−Bkζk−1
0
k(2α − k − 1)
2k + 1
−
k + 3
ζk+4
0
Φk+1
−
Ψk+2
ζk+4
0
(k + 3)(k + 2 + 2α)
2k + 5
+
Ψk
ζk+2
0
α − (k + 1)2
2k + 1
= 0. (8.15)
Likewise, from the condition σζζ(ζ0, ϑ) = 0,
we get the following equation,
−Ak+1k(k + 1)ζk−1
0

+Bk+2ζk+1
0
(k + 2){2 − α − (k − 1)(k + 2)}
2k + 5
+Bkζk−1
0
k(k + 1)(2α − k − 1)
(2k + 1)
−
(k + 2)(k + 3)
ζk+4
0
Φk+1
−Ψk+2
(k + 2)(k + 3)(k + 2 + 2α)
ζk+4
0 (2k + 5)
−Ψk
(k + 1){(k + 4)(k + 1) + α − 2}
ζk+2
0 (2k + 1)
= 0. (8.16)
Thus if we multiply (8.15) by k + 1 and add
(8.16), from the resulting equation we ﬁnd
Bk+2 =
(2k + 5)
ζk+1
0 {−α(2k + 3) + 2(k + 2)(k + 3)}
×
(k + 3)(2k + 3)
ζk+4
0
Φk+1
+Ψk+2
(2k + 3)(k + 3)(k + 2 + 2α)
ζk+4
0 (2k + 5)
+
Ψk
ζk+2
0
(k + 1)(k + 3) . (8.17)
0
Using the relation
Φk+1 = −
α − 1
k + 2
Ψk+2,
Bk+2 is
Bk+2 =
(2k + 5)
ζk+1
0 {−α(2k + 3) + 2(k + 2)(k + 3)}
× Ψk+2
(2k + 3)(k + 3)((k + 3)2 − α)
ζk+4
0 (2k + 5)(k + 2)
+
Ψk
ζk+2
0
(k + 1)(k + 3) . (8.18)
From (8.15) we have
Ak+3 =
Ψk
ζ2k+3
0
L(k) +
Ψk+2
ζ2k+5
0
M(k) +
Ψk+4
ζ2k+7
0
N(k),
where
L(k) = (k + 1)(k + 3)
(2α − k − 3)
H(k)
,
M(k) =
1
k + 2
(k + 3)
(2k + 3)(2α − k − 3)I(k)
(2k + 5)H(k)
+(k + 4)(k + 5)
G(k)
H(k + 2)
+
G(k + 2)
2k + 5
,
N(k) =
G(k)
H(k + 2)
(2k + 7) − 1
×
(k + 5)I(k + 2)
(2k + 9)(k + 2)(k + 3)
,
with
H(k) = −α(2k + 3) + 2(k + 2)(k + 3),
G(k) = 2−α−(k+1)(k+4), I(k) = −α+(k+3)2
.
Therefore using the formula in Theorem B of Ap-
pendix 4, we have
A =
∞
n=1
(−1)n
2nt2n−1
a2n
=
∞
n=1
(−1)n
2nt2n−1 1
δ2n(2n)!
∞
k=2n
k!Akδk
(k − 2n)!
=
∞
k=2
[k/2]
n=1
k!(−1)n
(2n − 1)!(k − 2n)!
t
δ
2n
1
t
×Akδk
=
∞
k=−1
fk(t)Ak+3 =
∞
k=−1
fk(t)
×
Ψk
ζ2k+3
0
L(k) +
Ψk+2
ζ2k+5
0
M(k) +
Ψk+4
ζ2k+7
0
N(k) .
(8.19)
In the above equation Ψk = 0, if k ≤ 0 and
fk(t) = δk+3
[(k+3)/2]
n=1
(k + 3)!(−1)n
(2n − 1)!(k + 3 − 2n)!
×
t
δ
2n
1
t
.
If we substitute the values of Ψk into (8.19),it
reduces to
A =
1
0
g(u)Ω1(t, u)du,
where
Ω1(t, u) =
∞
k=−1
fk(t)
hk(u)
ζ2k+3
0
L(k)
+
hk+2(u)
ζ2k+5
0
M(k) +
hk+4(u)
ζ2k+7
0
N(k) .

In the above equation hk(u) = 0, if k ≤ 0 and
hk(u) = δk
[(k−1)/2]
n=0
(−1)nk!
(2n + 1)!(k − 2n − 1)!
×
u
δ
2n+1
.
Also, using Theorem B of Appendix 4, we get
∞
n=0
(−1)n
b2n+1t2n+1
=
∞
n=0
(−1)n t2n+1
(2n + 1)!δ2n+1
×
∞
k=2n+1
k!Bkδk
(k − 2n − 1)!
=
∞
k=1
[ k−1
2
]
n=0
(−1)nk!
(2n + 1)!(k − 2n − 1)!
t
δ
2n+1
Bkδk
=
∞
k=−1
hk+2(t)Bk+2
=
∞
k=−1
hk+2(t)
2k + 5
ζk+1
0 H(k)
Ψk
ζk+2
0
(k + 1)(k + 3)
+
Ψk+2
ζk+4
0
(k + 3)(2k + 3)
2k + 5
I(k)
=
1
0
g(u)Ω2(t, u)du,
0
where
Ω2(t, u) =
∞
k=−1
hk+2(t)
2k + 5
ζk+1
0 H(k)
hk(u)
ζk+2
0
(k + 1)
×(k + 3) +
hk+2(u)
ζk+4
0
(k + 3)(2k + 3)
2k + 5
I(k) .
In the above equation hk(u) = 0, if k ≤ 0. Thus
(7.13) reduces to the following Fredholm integral
equation of the second kind
g(t) +
1
0
g(u)K(t, u)du = h(t),
where
K(t, u) = −
2
π
{Ω1(t, u) + αΩ2(t, u)}.
Appendix 1. Proof of (3.3). Since
ρ2
= (ζ cos ϑ − δ)2
+ (ζ sin ϑ)2
= ζ2
1 − 2 cos ϑ
δ
ζ
+
δ
ζ
2
,
1
ρ
=
1
ζ 1 − 2 cos ϑδ
ζ + (δ
ζ )2
=
∞
n=0
Pn(cos ϑ)
δn
ζn+1
,
Pn(cos θ)
ρn+1
=
(−1)n
n!
∂n
∂zn
1
ρ
=
∞
k=0
(−1)n
n!
∂n
∂zn
Pk(cos ϑ)
ζk+1
δk
=
∞
k=0
(−1)n
n!
∂n+k
∂zn+k
1
ζ
(−1)k
k!
δk
=
∞
k=0
n + k
k
Pn+k(cos ϑ)
ζn+k+1
δk
.
Appendix 2. Proof of (3.16).
2πρn
Pn(cos θ) =
π
−π
(z + ix cos u + iy sin u)n
du
=
π
−π
(−δ + Z + ix cos u + iy sin u)n
du
=
n
k=0
n!
(n − k)!k!
(−δ)n−k
×
π
−π
(Z + ix cos u + iy sin u)k
du
= 2π
n
k=0
n!
(n − k)!k!
(−δ)n−k
ζk
Pk(ϑ).
Appendix 3. Theorem A. If the equation
Bk =
k
i=0
k!
(k − i)!i!
ai,
is solved for ais, it will be written as
ai =
i
k=0
i!
(i − k)!k!
Bk(−1)i−k
.
Proof. We prove it by mathematical induction.
Suppose it is true for i, we will show that it is
also true for i + 1. Suppose Bi+1 is given by
Bi+1 =
i+1
k=0
(i + 1)!
(i + 1 − k)!k!
ak = ai+1

+
i
k=0
(i + 1)!
(i + 1 − k)!k!
ak.
Thus
ai+1 = Bi+1 −
i
k=0
(i + 1)!
(i + 1 − k)!k!
×
k
j=0
Bj
k!
j!(k − j)!
(−1)k−j
= Bi+1 −
i
j=0
Bj
(i + 1)!
j!
i
k=j
(−1)k−j
(i + 1 − k)!(k − j)!
.
(A.1)
The inner summation in the above equation can
written as, by changing the variable k − j = m
i
k=j
(−1)k−j
(i + 1 − k)!(k − j)!
=
i−j
m=0
(−1)m
(i + 1 − m − j)!m!
= −
(−1)i−j+1
(i − j + 1)!
,
0
since
i−j+1
m=0
(−1)m(i − j + 1)!
(i + 1 − m − j)!m!
= (1 − 1)i−j+1
= 0.
Then (A.1) is equal to
ai+1 = Bi+1 +
i
j=0
Bi(i + 1)!(−1)i+1−j
j!(i + 1 − j)!
=
i+1
j=0
Bi(i + 1)!(−1)i+1−j
j!(i + 1 − j)!
.
Appendix 4. Theorem B. If the equation
Bk =
∞
n=k
n
k
(−δ)n−k
an,
is solved for ans, it will be written as
an =
∞
k=n
k
n
δk−n
Bk.
Proof. Let
∞
n=0
an(−δ)n
xn
= f(x),
then
an(−δ)n
n! = f(n)
(0),
and
f(k)
(1) =
∞
n=k
an(−δ)n n!
(n − k)!
= k!Bk(−δ)k
.
From Taylor’s series
f(x) =
∞
k=0
f(k)(1)
k!
(x − 1)k
.
Thus
an(−δ)n
n! = f(n)
(0) =
∞
k=0
f(k)(1)
k!
dn
dxn
(x−1)k
x=0
=
∞
k=n
f(k)(1)
(k − n)!
(−1)k−n
=
∞
k=n
Bk(−1)k−n(−δ)kk!
(k − n)!
.
Therefore
an =
∞
k=n
k!
n!(k − n)!
Bkδk−n
.
References
[1] L. M. Keer, V.K. Luk, and J. M. Freedman,
Circumferential edge crack in a cylindrical cav-
ity, J.Appl.Mech. 44, 1977, pp.250-254
[2] L. M. Keer and H. A. Watts, Mixed boundary
value problems for a cylindrical shell, Int.J.Solids
Structures, 12, 1976, pp.723-729
[3] A. Atsumi and Y. Shindo, Axisymmet-
ric singular stresses in a thick-walled spherical
shell with an internal edge crack, J.Appl.Mech.
50,1983, pp.37-42.
[4] A. Atsumi and Y. Shindo, Peripheral edge
crack around spherical cavity under uniaxial ten-
sion ﬁeld, Int. J. Engng Sci. 21(3), 1983, pp.269-
278
[5] D.-S. Lee, Torsion of a long circular cylinder
having a spherical cavity with a peripheral edge
crack, Eur.J.Mech.A/Solids 21, 2002, pp.961-969

[6] D.-S. Lee, Tension of a long circular cylin-
der having a spherical cavity with a peripheral
edge crack , Int. J. Solids Structures, 40, 2003,
pp.2659-2671
[7] H. Wan, Q. Wang, and X. Zhang, Closed
form solution of stress intensity factors for
cracks emanating from surface semi-spherical
cavity in finite body with energy release
rate method, Appl.Math.Mech.Engl.Ed., 37(12),
2016, pp.1689-1706
[8] A.E. Green and W. Zerna, Theoretical Elas-
ticity. New York, Oxford Univ. Press, 1975.
[9] E.T. Whittaker and G.N. Watson, A Course
of Modern Analysis, Cambridge Univ. Press,
Cambridge, 1973. 0
[10] A. Erdélyi, W. Magnus, F. Oberhettinger,
and F.G. Tricomi, Tables of Integral Transforms
Vol.I, New York, McGraw-Hill, 1954
[11] F. Erdogan, G.D. Gupta, and T.S. Cook,
Numerical solution of singular integral equation.
Methods of analysis and solutions of crack prob-
lems, Noordhoff International Publishing, 1973,
pp.368-425
[12] K.N. Srivastava and J.P. Dwivedi, The ef-
fect of a penny-shaped crack on the distribution
of stress in an elastic sphere, Int.J.Engng. sci.,
9, 1971, pp.399-420,
[13] Ranjit S.Dhaliwal, Jon G. Rokne, and Brij
M. Singh, Penny-shaped crack in a sphere em-
bedded in an infinite medium, Int.J.Engng Sci.
17, 1979, pp.259-269
[14] R.P.Srivastav and Prem Narain, Some mixed
boundary value problems for regions with spher-
ical boundary, Journal of Math. Mech., 14, 1965,
pp.613-627

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